If {tan ^{ - 1}}(x - 1) + {tan ^{ - 1}}x + {t | vedclass

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(D) Given equation: ${	an ^{ - 1}}(x - 1) + {	an ^{ - 1}}x + {	an ^{ - 1}}(x + 1) = {	an ^{ - 1}}3x$ Rearranging the terms: ${	an ^{ - 1}}(x - 1)

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$ 0, \pm \frac{1}{2} $

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(D) Given equation: ${	an ^{ - 1}}(x - 1) + {	an ^{ - 1}}x + {	an ^{ - 1}}(x + 1) = {	an ^{ - 1}}3x$ Rearranging the terms: ${	an ^{ - 1}}(x - 1) + {	an ^{ - 1}}(x + 1) = {	an ^{ - 1}}3x - {	an ^{ - 1}}x$ Applying the formula ${	an ^{ - 1}}A + {	an ^{ - 1}}B = {	an ^{ - 1}}\left( \frac{A+B}{1-AB} ight)$ and ${	an ^{ - 1}}A - {	an ^{ - 1}}B = {	an ^{ - 1}}\left( \frac{A-B}{1+AB} ight)$: ${	an ^{ - 1}}\left( \frac{(x-1)+(x+1)}{1-(x-1)(x+1)} ight) = {	an ^{ - 1}}\left( \frac{3x-x}{1+(3x)(x)} ight)$ ${	an ^{ - 1}}\left( \frac{2x}{1-(x^2-1)} ight) = {	an ^{ - 1}}\left( \frac{2x}{1+3x^2} ight)$ This implies: $\frac{2x}{2-x^2} = \frac{2x}{1+3x^2}$ Case $1$: $2x = 0 \Rightarrow x = 0$ Case $2$: $\frac{1}{2-x^2} = \frac{1}{1+3x^2}$ $1 + 3x^2 = 2 - x^2$ $4x^2 = 1 \Rightarrow x^2 = \frac{1}{4} \Rightarrow x = \pm \frac{1}{2}$ Thus,the solutions are $x = 0, \pm \frac{1}{2}$.

If ${\tan ^{ - 1}}(x - 1) + {\tan ^{ - 1}}x + {\tan ^{ - 1}}(x + 1) = {\tan ^{ - 1}}3x$,then $x =$

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